Question

An aqueous 0.300 M glucose solution is prepared with a total volume of 0.150 L. The molecular weight of

Answer 1

Answer:

We need 8.11 grams of glucose for this solution

Explanation:

Step 1: Data given

Molarity of the glucose solution = 0.300 M

Total volume = 0.150 L

The molecular weight of glucose = 180.16 g/mol

Step 2: Calculate moles of glucose in the solution

Moles glucose = molarity solution * volume

Moles glucose = 0.300 M * 0.150 L

Moles glucose = 0.045 moles glucose

Step 3: Calculate mass of glucose

MAss glucose = moles glucose* molecular weight of glucose

MAss glucose = 0.045 moles * 180.16 g/mol

MAss glucose = 8.11 grams

We need 8.11 grams of glucose for this solution